Solution Manual For Introductory Circuit Analysis, 12th Edition

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Chapter 11Chapter 11.2.3.4.=td=h1min60min1s60ft5,280mi1s10ft000,20=1363.64 mph5.4 minmin60h1=0.067 h=h1.067mi31td=29.05 mph6.100 ydsft5,280mi1yd1ft3= 0.0568 mi60 mi1 h1 minh60 min60 s   = 0.0167 mi/st=0.0568 mi0.0167 mi/sd =3.40 s7.a.s60min1min60h1mift280,5hmi95=139.33 ft/sb.t =ft/s139.33ft60d=0.431 sc.=ft5,280mi1h1min60min1s60s1ft60td=40.91 mph8.9.10.11.MKS, CGS,C =)36(95)3268(95)32F(95=20SI: K = 273.15 +C = 273.15 + 20 =293.1512.J1lb-ft0.7378J1000=737.8 ft-lbs

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Chapter 1213.0.5 ydin.1cm54.2ft1in.12yd1ft3=45.72 cm14.a.°F = 2(°C) + 30° = 40° + 30° =70°b.°F =99( C)32(20 )3255  =68°c.very closed.30°C90°F vs. 86°5°C40°F vs 41°15.a.14.6b.56.0c.1,046.1d.116= 0.0625 =0.1e.3.14159 =3.116.a.14.60b.56.04c.1,046.06d.116= 0.0625 =0.06e.3.14159 =3.1417.a.14.603b.56.042c.1,046.060d.116= 0.0625 =0.063e.3.14159 =3.14218.a.104b.106c.103d.103e.100f.10119.a.15103b.5103c.2.4106d.60103e.4.02104f.2101020.a.4.2103+ 48.0103= 52.2103=5.22104b.90103+ 360103= 450103=4.50105c.501056105= 44105=4.4104d.1.2103+ 0.051030.4103= 0.85103=85021.a.(102)(103) = 105=100103b.(102)(103) = 101=10c.(103)(106) =1109d.(102)(105) =1103e.(106)(10106) =10f.(104)(108)(1028) =11024

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Chapter 1322.a.(50103)(2103) = 100100=100b.(2.2103)(2103) = 4.4100=4.4c.(82106)(1.2106) =98.4d.(30104)(4103)(7108) = 840101=8.4010323.a.102/104= 102=10103b.102/103= 105=10106c.104/103= 107=10106d.107/102=1.0109e.1038/104=1.01042f.210/100= 101/102=110324.a.(2103)/(8105) = 0.25108=2.50107b.(4103)/(4106) = 4/4109=1109c.(22105)/(5105) = 22/5100=4.4d.(781018)/(4106) =1.95102525.a.(102)3=1.0106b.(104)1/2=10.0103c.(104)8=100.01030d.(107)9=1.0106326.a.(2102)2=4104b.(5103)3=125109c.(4103)(3103)2= (4103)(9104) = 36101=360d.((2103)(0.8104)(0.003105))3= (4.8103)3= (4.8)3(103)3= 110.6109=1.11101127.a.(103)2=1.0106b.34210)10)(10(= 102/103=1.0105c.444264223101010)10)(10(10)10()10(=1.0108d.44310)10)(10(= 107/104=1.01011e.(1104)3(102)/106= (1012)(102)/106= 1010/106=1.01016f. 101)10)(10(110)10()10)(10(34322322=1.010128.a.4222103)10()103(= (9104)(102)/(3104) = (9106)/(3104) = 3102=300b.383241081016)20()104(= 2105c.4822241041036)102()106(=9.01012

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Chapter 14d.5253/16102103102)1027(= 1.5107=150.0109e.4842642231021048102)103)(1016(102)103()104(=24.01012f.(16106)1/2(105)5(2102) = (4103)(1025)(2102) = 81020=800.01018g.21/ 23322452(310)1.6010(210 )(810)(710)  9421/ 2105161010(2710)(2.5610 )(1610)4910(69.1210)(410)276.481049104910= 5.64104=56.410329.Scientific:a.2.05 × 101b.5.04 × 104c.6.74 × 104d.4.60 × 102Engineering:a.20.46 × 100b.50.42 × 103c.674.00 × 106d.46.00 × 10330.Scientifica.5.0 × 102b.4.5 × 101c.1/32 = 0.03125 =3.125 × 102d.3.14159 =3.142 × 100Engineering:a.50.0 × 103b.0.045 × 103c.31.25 × 103d.3.142 × 100

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Chapter 15+231.a.6104= 0.0610+6=0.06 × 10+633b.0.4103= 400106=400 × 106+32+3+3c.50105= 5000103= 5106= 0.005109=0.005 × 109+233+433d.12107= 0.0012103= 1.2106= 1200109=1200 × 1094+3+3332.a.0.05100s = 50103s =50 ms+3+3b.2000106s = 2103s =2 ms33c.0.04103s = 40106s =40s+3+6d.84001012s0.0084106s =0.0084s6

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Chapter 16+3increase by 3100e.100103103m = 0.1103m =0.1 km333.a.1.5 minmin1s60=90 sb.2 × 102hmin1s60h1min60=72 sc.0.05 ss10s16= 0.05106s =50103sd.0.16 mm10mm13= 0.16103mm =160 mme.1.2107ss10ns19= 1.2102ns =120 nsf.4108sh24day1min60h1s60min1=4629.63 days34.a.80103mm1cm100= 8000103cm =8 cmb.60 cmm1000km1cm100m1=60105kmc.12 × 103m61m10m= 12103× 10+6m =12 × 103md.60 cm21 m1 m100 cm100 cm   =60104m235.a.100 in.in.39.37m1=2.54 mb.4 ftin.39.37m1ft1in.12=1.22 m

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Chapter 17c.6 lblb1N45.4=26.7 Nd.60103dynesN4.45lb1dynes10N15=0.13 lbe.150,000 cmin.12ft1cm2.54in.1=4921.26 ftf.0.002 miin.39.37m1ft1in.12mi1ft5280=3.22 m36.5280 ft,5280 ftft3yd1=1760 yds5280 ftin.39.37m1ft1in.12= 1609.35 m, 1.61 km37.60 mi5280 ft12 in.1 m1 h1 minh1 mi1 ft39.37 in.60 min60 s=26.82 m/s38.10 kmft5280mi1in.12ft1m1in.39.37km1m1000= 6.214 mi=min6.5mi1,t=min6.5mi1mi6.214d=40.39 min39.100 ydsft1in.12yd1ft3= 3600 in3600 quarters40.60 mph:t=500 mi60 mi/hd = 8.33 h =8 h:19.8 min75 mph:t=500 mi75 mi/hd = 6.67 h =6h:40.2 mindifference =1h:28.6 minutes41.d=t=cm100m1min1s60h1min60h016.0scm600=345.6 m

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Chapter 1842.d= 86 storiesft129step1storyft14= 1605 steps=seconds60minute1seconds5.802secondsteps2steps1605dttd=13.38 minutes43.d= (86 stories)ft5,280mile1ft1204storyft14= 0.228 milesmin10.22 minmile0.228 miles=44.82 min/mile44.minuteft10561mileft5,280min5mile1milemin5,distance = 86 storiesstoryft14= 1204 ft=minft1056ft1204dttd=1.14 minutes45.a.J1054.35Btu1J5=4.74103Btub.24 ouncesgallons172.264m1ounces128gallon13=7.1104m3c.1.4 daysday1s400,86=1.21105sd.1 m3gallon1pints8m1gallons264.1723=2113.38 pints46.6(4 × 2 + 8) =9647.(42 + 6/5)/3 =14.448.22253= 5.04449.MODE = DEGREES: cos 21.87=0.92850.MODE = DEGREES: tan1(3/4) =36.8751.2400 /(610 / 5)= 7.071

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Chapter 1952.20510653.1.20101254.6.667106+ 0.5106=7.17106

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Chapter 210Chapter 21.2.a.F=29221m)(1C)C)(21)(109(rQQk=18109Nb.F=k221rQQ=29m)3(C2C)1)(109(=2109Nc.F=29221m)(10C)C)(21)(109(rQQk=0.18109Nd.Exponentially,3110 m1 mrr= 10 while91921810N0.1810NFF=1003.a.r= 1 mi:1 miin.37.39m1ft1in.12mi1ft5280= 1609.35 mF=96-6312226(910 )(810C)(4010C)288010(1609.35 m)2.5910kQ Qr=1.11Nb.r=10 ft:12 in.1 m10 ft1 ft39.37 in.   = 3.05 mF=3312222880102880109.30(3.05 m)kQ Qr=0.31 Nc.in.39.37m116in.1= 1.59 mmF=3331223262880102880101138.3410(1.5910m)2.5310kQ QrN=1138.34 kN4.5.Q1=Q2=Q;222221 11 112222122;FrF rkQkQkFQFkkrrrand2F21122rFr6.F=426921221106.3)1020)(109(FQkQrrQkQ=10 mm

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Chapter 2117.F=21221221m)2(8.1QkQQkQrQkQ= 4(1.8) = 7.2a.F=2221)10(2.7rQkQ=72 mNb.Q1/Q2= 1/2Q2= 2Q17.2 =kQ1Q2= (9109)(Q1)(2Q1) = 9109212Q9121910182.710182.7QQ=20CQ2= 2Q1= 2(2105C) =40C8.V=1.2 J20CWQ=60 kV9.W=VQ= (60 V)(8 mC) =0.48 J10.Q=120J20 mVWV= 6 mC186.24210electrons1 C= 37.45 × 1015electrons11.Q=V9J72VW=8 C12.a.W=QV= (1 × 1012electrons)(40 V) =40 × 1012eVb.40 × 1012eV181 C6.24210electrons=6.41μJ13.I=12 mC2.8 sQt=4.29 mA14.I=312 C(2)(60 s)Qt=2.60 A15.Q=It= (40 mA)(0.8)(60 s) =1.92 C16.Q=It= (250 mA)(1.2)(60 s) =18.0 C17.t=mA2mC6IQ=3 s18.21.8471018electronselectrons106.242C118= 3.5 CI=s12C3.5tQ= 0.29 A

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Chapter 21219.Q=It= (4 mA)(90 s) = 360 mC360 mCC1electrons10242.618=2.251018electrons20.I=s)(1.2)(60C86tQ= 1.194 A > 1 A(yes)21.0.841016electronselectrons106.242C118= 1.346 mCI=ms60mC346.1tQ=22.43 mA22.a.Q=It= (2 mA)(0.01s) = 21011C21011C186.24210electrons1 ¢1 Celectron = 1.25108¢= $1.25106=1.25 millionb.Q=It= (100A)(1.5 ns) = 1.51013C1.51013Celectron1$C1electrons10242.618=0.94 million(a) > (b)23.Q=It= (200103A)(30 s) = 6 CV=C6J40QW=6.67 V24.Q=It=min)5.0(minC420= 210 CV=C210J742QW=3.53 V25.Q=V24J0.4VW= 0.0167 CI=30.0167 C510sQt=3.34 A26.I=h40Ah200(hours)ratingAht=5 A27.Ah = (0.8 A)(75 h) =60.0 Ah28.t(hours) =A1.28Ah32ratingAhI=25 h

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Chapter 21329.40 Ah(for 1 h):W1=VQ=VIt= (12 V)(40 A)(1 h)min1s60h1min60= 1.728106J60 Ah(for 1 h):W2= (12 V)(60 A)(1 h)min1s60h1min60= 2.592106JRatioW2/W1= 1.5 or 50% more energy available with 60 Ah rating.For 60 s discharge: 40 Ah =It=Imin60h1s60min1s60=I(16.67103h)andI=h1016.67Ah403-=2400 A60 Ah =It=Imin60h1s60min1s60=I(16.67103h)andI=h1016.67Ah603-=3600 AI2/I1= 1.5 or 50 % more starting current available at 60 Ah30.0.75(18 Ah) = 13.5 Ah250 mA31.(18 Ah15.5 Ah)/18 Ah × 100% =13.89%32.At 100 mA, discharge time120 H; At 25 mA, discharge time425 h;300 h more at 25 mA33.I=3 Ah6.0 h= 500 mAQ=It= (500 mA)(6 h)min1s60h1min60= 10.80 kCW=QV= (10.8 kC)(12 V)129.6 kJ34.35.36.37.a.0.5 in2.54 cm1 in= 1.27 cm1.27 cm30 kVcm=38.1 kVb.1.27 cm270 kVcm=342.9 kVc.342.9 kV:38.1 kV =9:1

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Chapter 21438.39.40.41.

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Solution Manual For Introductory Circuit Analysis, 12th Edition - Page 16 preview image

Chapter 315Chapter 31.a.0.5 in. =500 milsb.0.02 in.in.1mils1000=20 milsc.41in. = 0.25 in.in.1mils1000=250 milsd.10 mm = 10 × 103m39.37 in1000 mils1 m1 in  =393.7 milse.0.01 ftin.1mils10ft1in.123=120 milsf.0.1 cmin.1mils1000cm2.54in.1=39.37 mils2.a.ACM= (30 mils)2=900 CMb.0.016 in. = 16 mils,ACM= (16 mils)2=256 CMc.18= 0.125" = 125 mils,ACM= (125 mils)2=15.63103CMd.1 cmin.1mils1000cm2.54in.1= 393.7 mils,ACM= (393.7 mils)2=155103CMe.0.02 ftin.1mils1000ft1in.12= 240 mils,ACM= (240 mils)2=57.60103CMf.4 × 103m39.37 in.1000 mils1 m1 in   =157.48 mils,ACM= (157.48 mils)2=24.8 × 103CM3.ACM= (dmils)2dmils=CMAa.d=CM1600= 40 mils =0.04 in.b.d=820 CM= 28.64 mils =0.029 in.c.d=CM40,000= 200 mils =0.2 in."

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